$x(1)=1$ and $x(n)=x(n-1)+(\frac1n)$ for $n>1$
So far, I tried looking at $|x(n+1)-x(n)|$ to get $|x(n)+(\frac1n)-x(n)|$.
I also know that $\frac1n$ diverges. I am stuck on what to do next.
$x(1)=1$ and $x(n)=x(n-1)+(\frac1n)$ for $n>1$
So far, I tried looking at $|x(n+1)-x(n)|$ to get $|x(n)+(\frac1n)-x(n)|$.
I also know that $\frac1n$ diverges. I am stuck on what to do next.
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You have the two things. Just put them together. $|x_n - x_m| = |\sum_{i=n}^m 1/i |$ (why is this true?). Thus, it is not true that for all $n,m > N$, that quantity can be made small. Fix $n$, make $m$ as large as is necessary to exceed $\epsilon$.
I leave it to you to write down a detailed proof that would meet the standards of the question.